Difference between revisions of "2009 AMC 12B Problems/Problem 9"
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=== Solution 4 === | === Solution 4 === | ||
− | WLOG, let the coordinates of <math>C</math> be <math>(3,4)</math> , or any coordinate, for that matter. Applying the shoelace formula, we get the area as \boxed 6 | + | WLOG, let the coordinates of <math>C</math> be <math>(3,4)</math> , or any coordinate, for that matter. Applying the shoelace formula, we get the area as <math>\boxed 6</math>$ . |
== See also == | == See also == |
Revision as of 21:18, 27 April 2020
Contents
Problem
Triangle has vertices , , and , where is on the line . What is the area of ?
Solution
Solution 1
Because the line is parallel to , the area of is independent of the location of on that line. Therefore it may be assumed that is . In that case the triangle has base and altitude , so its area is .
Solution 2
The base of the triangle is . Its altitude is the distance between the point and the parallel line , which is . Therefore its area is . The answer is .
Solution 3
By Shoelace, our area is: We know so we get:
Solution 4
WLOG, let the coordinates of be , or any coordinate, for that matter. Applying the shoelace formula, we get the area as $ .
See also
2009 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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